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enter image description here

How to solve the above question?

I cannot apply Kirchoff's current equation because of the presence of the dependent voltage source

Also, I am not able to apply supernode because of the 2 kOhm resistor in series. How should I go about it? Please help.

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  • \$\begingroup\$ What current is going through the 1k resistor? \$\endgroup\$
    – Transistor
    Commented May 10, 2020 at 12:03
  • \$\begingroup\$ @Transistor \$ v_2/1000 \$ Amperes. \$\endgroup\$
    – Soumee
    Commented May 10, 2020 at 12:05
  • \$\begingroup\$ You can get the numerical value just by looking at that loop. What's the current on the other side of the loop? Then what's the current through the 1k resistor? \$\endgroup\$
    – Transistor
    Commented May 10, 2020 at 12:26
  • \$\begingroup\$ @Transistor 3 mAmperes \$\endgroup\$
    – Soumee
    Commented May 10, 2020 at 12:40
  • \$\begingroup\$ So, can you now work out \$ v_2 \$? \$\endgroup\$
    – Transistor
    Commented May 10, 2020 at 12:47

2 Answers 2

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Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_3}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3} \end{cases}\tag2 $$

We also see that \$\text{V}_2-\text{V}_3=\text{n}\cdot\text{V}_2\$.

Now, using those equations your problem can be solved.

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Everything is in series with the current source, meaning the voltage across the 1K resistor must be 1000*0.003, from there you solve the rest of the circuit

The step after this, as you know the V2 node would be how that current source is split between the 2K + voltage source and the 4K Resistor, if you need to, you can imagine it as 2 voltage sources with an ESR, both in parallel with each other.

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